Question #2e6dd
1 Answer
Explanation:
Start by comparing the final temperature of the water + aluminium sample with the initial temperatures of the two substances.
Notice that aluminium goes from an initial temperature of
On the other hand, the water goes from an initial temperature of
The idea here is that the heat lost by the metal will be equal to the heat gained by the water.
color(blue)(-q_"metal" = q_"water")" " " "color(orange)("(*)")
The Minus sign is used here because heat lost will always carry a negative sign.
Now, in order to determine how much heat was lost / gained, you can use the equation
color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" " , where
Water's specific heat can be found here
c_"water" = "4.18 J g"^(-1)""^@"C"^(-1)
https://en.wikipedia.org/wiki/Heat_capacity#Mass_heat_capacity_of_building_materials
So, you know the mass of water, its specific heat, and its change in temperature
DeltaT_"water" = 26.1^@"C" - 23.7^@"C" = 2.40^@"C"
so plug in your values and find how much heat was absorbed by the water
q_"water" = 21.9 color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * 2.4color(red)(cancel(color(black)(""^@"C")))
q_"water" = "219.7 J"
This means that the aluminium must have lost the exact same amount of heat. Use equation
q_"metal" = -"219.7 J"
You know that you have
q_"metal" = m_"metal" * c_"metal" * DeltaT_"metal"
The change in temperature for the metal will be
DeltaT_"metal" = 26.1^@"C" - 100.0^@"C" = -73.9^@"C"
Rearrange the equation to solve for
m_"metal" = q_"metal"/(c_"metal" * DeltaT_"metal")
m_"metal" = (color(blue)(cancel(color(black)(-)))219.7color(red)(cancel(color(black)("J"))))/(0.900color(red)(cancel(color(black)("J")))/("g" * color(red)(cancel(color(black)(""^@"C")))) * (color(blue)(cancel(color(black)(-)))73.9color(red)(cancel(color(black)(""^@"C"))))) = color(green)(|bar(ul(color(white)(a/a)"3.30 g"color(white)(a/a)|)))
The answer is rounded to three sig figs.